从多个表插入到其他表中

时间：2015-06-27 20:00:21

标签： mysql sql

这是我尝试加入的七个表的表结构：

-- tables: en, fr, de, zh_cn, es, ru, pt_br
`geoname_id` INT (11),
`continent_code` VARCHAR (200),
`continent_name` VARCHAR (200),
`country_iso_code` VARCHAR (200),
`country_name` VARCHAR (200),
`subdivision_1_name` VARCHAR (200),
`subdivision_2_name` VARCHAR (200),
`city_name` VARCHAR (200),
`time_zone` VARCHAR (200)

这是新的表结构，其中将存储所有数据：

CREATE TABLE `geo_lists` (
    `city_id` int (11), -- en.geoname_id (same for all 7 tables)
    `continent_code` varchar (2), -- en.continent_code (same for all 7 tables)
    `continent_name` varchar (200), -- en.continent_name (just in english)
    `country_code` varchar (2), -- en.country_iso_code (same for all 7 tables)
    `en_country_name` varchar (200), -- en.country_name
    `fr_country_name` varchar (200), -- fr.country_name
    `de_country_name` varchar (200), -- de.country_name
    `zh_country_name` varchar (200), -- zh_cn.country_name
    `es_country_name` varchar (200), -- es.country_name
    `ru_country_name` varchar (200), -- ru.country_name
    `pt_country_name` varchar (200), -- pt_br.country_name
    `en_state_name` varchar (200), -- en.subdivision_1_name
    `fr_state_name` varchar (200), -- fr.subdivision_1_name
    `de_state_name` varchar (200), -- de.subdivision_1_name
    `zh_state_name` varchar (200), -- zh_cn.subdivision_1_name
    `es_state_name` varchar (200), -- es.subdivision_1_name
    `ru_state_name` varchar (200), -- ru.subdivision_1_name
    `pt_state_name` varchar (200), -- pt_br.subdivision_1_name
    `en_province_name` varchar (200), -- en.subdivision_2_name
    `fr_province_name` varchar (200), -- fr.subdivision_2_name
    `de_province_name` varchar (200), -- de.subdivision_2_name
    `zh_province_name` varchar (200), -- zh_cn.subdivision_2_name
    `es_province_name` varchar (200), -- es.subdivision_2_name
    `ru_province_name` varchar (200), -- ru.subdivision_2_name
    `pt_province_name` varchar (200), -- pt_br.subdivision_2_name
    `en_city_name` varchar (200), -- en.city_name
    `fr_city_name` varchar (200), -- fr.city_name
    `de_city_name` varchar (200), -- de.city_name
    `zh_city_name` varchar (200), -- zh_cn.city_name
    `es_city_name` varchar (200), -- es.city_name
    `ru_city_name` varchar (200), -- ru.city_name
    `pt_city_name` varchar (200), -- pt_br.city_name
    `time_zone` varchar (30) -- en.time_zone (same for all 7 tables)
);

我想加入所有这些，使用区域设置（语言）代码作为列名的前缀。

5 个答案:

答案 0 :(得分：3)

哦！ @GabrielBlanca你是对的，在这种情况下尝试这个查询，让我知道它是否有效。您可以复制并粘贴：

insert into geo_lists 
-- columns
(city_id, continent_code, continent_name, country_code, time_zone, 
en_country_name,
fr_country_name,
de_country_name,
zh_country_name,
es_country_name,
ru_country_name,
pt_country_name,

en_state_name,
fr_state_name,
de_state_name,
zh_state_name,
es_state_name,
ru_state_name,
pt_state_name,

en_province_name,
fr_province_name,
de_province_name,
zh_province_name,
es_province_name,
ru_province_name,
pt_province_name, 

en_city_name,
fr_city_name,
de_city_name,
zh_city_name,
es_city_name,
ru_city_name,
pt_city_name) 

-- end columns

select 
en.city_id, en.continent_code, en.continent_name, en.country_code, en.time_zone,
en.country_name as en_country_name, 
fr.country_name as fr_country_name,
de.country_name as de_country_name,
zh.country_name as zh_country_name,
es.country_name as es_country_name,
ru.country_name as ru_country_name,
pt.country_name as pt_country_name,

en.state_name as en_state_name,
fr.state_name as fr_state_name,
de.state_name as de_state_name,
zh.state_name as zh_state_name, 
es.state_name as es_state_name, 
ru.state_name as ru_state_name, 
pt.state_name as pt_state_name,

en.province_name as en_province_name, 
fr.province_name as fr_province_name, 
de.province_name as de_province_name,
zh.province_name as zh_province_name, 
es.province_name as es_province_name, 
ru.province_name as ru_province_name, 
pt.province_name as pt_province_name,

en.city_name as en_city_name, 
fr.city_name as fr_city_name, 
de.city_name as de_city_name, 
zh.city_name as zh_city_name,
es.city_name as es_city_name, 
ru.city_name as ru_city_name, 
pt.city_name as pt_city_name

from en, fr, de, zh_cn, es, ru, pt_br 

where en.city_id = fr.city_id 
and fr.city_id = de.city_id 
and de.city_id = zh_cn.city_id 
and zh_cn.city_id = es.city_id 
and es.city_id = ru.city_id 
and ru.city_id = pt_br.city_id

答案 1 :(得分：3)

好的Gabriel我们现在在哪里优化。

  1. 您在数据加载方面取得了哪些成就

  2. 带索引的表结构

  3. 行数

    4. 我会在这里提出一个答案，因为我将你从一个你愿意删除的重复问题中拖回来（thx）

答案 2 :(得分：2)

我还看到了有关查询优化的重复问题。我找到了解决方案。问题是您的表没有索引。只是做：

ALTER TABLE en
ADD PRIMARY KEY(geoname_id);

ALTER TABLE fr
ADD PRIMARY KEY(geoname_id);

ALTER TABLE de
ADD PRIMARY KEY(geoname_id);

ALTER TABLE zh_cn
ADD PRIMARY KEY(geoname_id);

ALTER TABLE es
ADD PRIMARY KEY(geoname_id);

ALTER TABLE ru
ADD PRIMARY KEY(geoname_id);

ALTER TABLE pt_br
ADD PRIMARY KEY(geoname_id);

然后运行：

DROP TABLE IF EXISTS geo_lists;

CREATE TABLE `geo_lists` (
    `city_id` int (11), -- en.geoname_id (same for all 7 tables)
    `continent_code` varchar (2), -- en.continent_code (same for all 7 tables)
    `continent_name` varchar (200), -- en.continent_name (just in english)
    `country_code` varchar (2), -- en.country_iso_code (same for all 7 tables)
    `en_country_name` varchar (200), -- en.country_name
    `fr_country_name` varchar (200), -- fr.country_name
    `de_country_name` varchar (200), -- de.country_name
    `zh_country_name` varchar (200), -- zh_cn.country_name
    `es_country_name` varchar (200), -- es.country_name
    `ru_country_name` varchar (200), -- ru.country_name
    `pt_country_name` varchar (200), -- pt_br.country_name
    `en_state_name` varchar (200), -- en.subdivision_1_name
    `fr_state_name` varchar (200), -- fr.subdivision_1_name
    `de_state_name` varchar (200), -- de.subdivision_1_name
    `zh_state_name` varchar (200), -- zh_cn.subdivision_1_name
    `es_state_name` varchar (200), -- es.subdivision_1_name
    `ru_state_name` varchar (200), -- ru.subdivision_1_name
    `pt_state_name` varchar (200), -- pt_br.subdivision_1_name
    `en_province_name` varchar (200), -- en.subdivision_2_name
    `fr_province_name` varchar (200), -- fr.subdivision_2_name
    `de_province_name` varchar (200), -- de.subdivision_2_name
    `zh_province_name` varchar (200), -- zh_cn.subdivision_2_name
    `es_province_name` varchar (200), -- es.subdivision_2_name
    `ru_province_name` varchar (200), -- ru.subdivision_2_name
    `pt_province_name` varchar (200), -- pt_br.subdivision_2_name
    `en_city_name` varchar (200), -- en.city_name
    `fr_city_name` varchar (200), -- fr.city_name
    `de_city_name` varchar (200), -- de.city_name
    `zh_city_name` varchar (200), -- zh_cn.city_name
    `es_city_name` varchar (200), -- es.city_name
    `ru_city_name` varchar (200), -- ru.city_name
    `pt_city_name` varchar (200), -- pt_br.city_name
    `time_zone` varchar (30) -- en.time_zone (same for all 7 tables)
);

INSERT INTO geo_lists
SELECT
    en.geoname_id,
    en.continent_code,
    en.continent_name,
    en.country_iso_code,
    en.country_name AS en_country_name,
    fr.country_name AS fr_country_name,
    de.country_name AS de_country_name,
    zh_cn.country_name AS zh_cn_country_name,
    es.country_name AS es_country_name,
    ru.country_name AS ru_country_name,
    pt_br.country_name AS pt_br_country_name,
    en.subdivision_1_name AS en_subdivision_1_name,
    fr.subdivision_1_name AS fr_subdivision_1_name,
    de.subdivision_1_name AS de_subdivision_1_name,
    zh_cn.subdivision_1_name AS zh_cn_subdivision_1_name,
    es.subdivision_1_name AS es_subdivision_1_name,
    ru.subdivision_1_name AS ru_subdivision_1_name, 
    pt_br.subdivision_1_name AS pt_br_subdivision_1_name,
    en.subdivision_2_name AS en_subdivision_2_name,
    fr.subdivision_2_name AS fr_subdivision_2_name,
    de.subdivision_2_name AS de_subdivision_2_name,
    zh_cn.subdivision_2_name AS zh_cn_subdivision_2_name,
    es.subdivision_2_name AS es_subdivision_2_name,
    ru.subdivision_2_name AS ru_subdivision_2_name,
    pt_br.subdivision_2_name AS pt_br_subdivision_2_name,
    en.city_name AS en_city_name,
    fr.city_name AS fr_city_name,
    de.city_name AS de_city_name,
    zh_cn.city_name AS zh_cn_city_name,
    es.city_name AS es_city_name,
    ru.city_name AS ru_city_name,
    pt_br.city_name AS pt_br_city_name,
    en.time_zone
FROM en
INNER JOIN fr USING(geoname_id)
INNER JOIN de USING(geoname_id)
INNER JOIN zh_cn USING(geoname_id)
INNER JOIN es USING(geoname_id)
INNER JOIN ru USING(geoname_id)
INNER JOIN pt_br USING(geoname_id);

在我的电脑上，查询时间是8.3秒。祝你好运！

答案 3 :(得分：1)

尝试这样的事情：

 INSERT INTO new_table (continent_code, subdivision_1_name) 
 SELECT en.continent_code , en.subdivision_1_name 
 FROM en

修改

 INSERT INTO new (cz_val,cz_value,en_val,en_value)
 SELECT cz.val, cz.value, en.val, en.value
 FROM cz
 INNER JOIN en ON en.id = cz.id
 GROUP BY cz.id

答案 4 :(得分：1)

试试这个：

INSERT INTO geo_lists
SELECT * FROM en UNION 
SELECT * FROM fr UNION 
SELECT * FROM de UNION 
SELECT * FROM zh_cn UNION
SELECT * FROM es UNION 
SELECT * FROM ru UNION 
SELECT * FROM pt_br UNION
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